A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the work done by the porter in lowering the suitcase. (take g =9.8 $m s^{- 2})$

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+627.2J

784.0J

-627.2J

-62720.0J

answer is B.

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Detailed Solution

From work-energy theorem
$\begin{array}{l} W_{porte r} + W_{mgE} = ΔK \cdot E . = 0 \\ (∵ velocity cons tant) \\ or, W_{porter} = - W_{mg} = - mgh \end{array}$
$∴ W_{por t e r} = - 80 \times 9 .8 \times \frac{80}{100} = - 627 .2 J$

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